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If an elliptic curve over $\mathbb{Q}$ has Mordell-Weil group of rank $r$, then its canonical height of an elliptic curve is a positive-definite quadratic form on the free lattice. Consequently

The canonical height pairing is $⟨P,Q⟩=(\hat{h}(P+Q)-\hat{h}(P)-\hat{h}(Q))/2$. The height parallelogram identity makes it bilinear. It descends through torsion and is positive definite on the free part of the Mordell-Weil group. The Cauchy-Schwarz inequality gives $\hat{h}(P-Q)\le 2\hat{h}(P)+2\hat{h}(Q)$, useful for height descent.

A positive integer is a congruent number when it is the area of a right triangle with rational side lengths. For square-free $D$, this is equivalent to the congruent number elliptic curve having a Mordell-Weil group of positive rank.

There is no nonconstant four-term arithmetic progression of rational square numbers. If $r^2,s^2,1,t^2$ formed one, then $(-2s^2,2rst)$ would lie on $y^2=x(x+1)(x+4)$; its explicitly known Mordell-Weil group forces $s^2=1$ and the common difference to vanish.

Suppose an elliptic curve $E/K$ has all four geometric 2-torsion points rational. For $P\in E(K)$ choose $Q\in E(\overline{K})$ with $2Q=P$. The function $c_P(\sigma )=\sigma (Q)-Q$ takes values in $E[2]$ and is a group homomorphism from the absolute Galois group to $E[2]$: its usual cocycle law becomes additivity because the action on $E[2]$ is trivial. Replacing $Q$ by another half, or $P$ by $P+2R$ with $R\in E(K)$, leaves it unchanged. Adding chosen halves proves additivity in $P$. Its kernel in $E(K)/2E(K)$ is zero, since $c_P=0$ precisely when the chosen half is rational. This constructs the injective map directly, without assuming any theorem of two-descent on an elliptic curve or finite generation of the Mordell-Weil group.

A group homomorphism between abelian groups, conventionally written additively. Its kernel and image are subgroups, and the additive formula implies preservation of zero, negation and every integer multiple. Maps on Mordell-Weil groups induced by isogenies of elliptic curves and by good reduction are examples.

For $E:y^2=x(x+1)(x+4)$,The points $(-2,2)$ and $(-1,0)$ generate the two factors, and a two-descent on an elliptic curve proves that the Mordell-Weil group has rank zero.

The saturation of a subgroup $H$ of the Mordell-Weil group consists of all points with a positive multiple in $H$. Saturation includes the full torsion subgroup. For a full-rank $H$ containing that torsion, it equals $E(K)$, but the index of $H$ may exceed one. Checking that $\ell P\in H$ forces $P\in H$ for every prime divisor of a certified index bound proves that $H$ is already the whole group. A canonical height pairing proves independence of found points; saturation answers the separate question of whether a full-rank subgroup is the full group.

The Mordell-Weil theorem gives $E(\mathbb{Q})\cong E(\mathbb{Q}{)}_{\mathrm{tors}}\oplus {\mathbb{Z}}^r$. Move the rational point of order two to $(0,0)$ and clear denominators to obtain an integral Weierstrass equation of an elliptic curveThe two-isogeny formula gives $E^{\prime }:Y^2=X(X^2-2aX+a^2-4b)$ and dual isogenies $\varphi :E\to E^{\prime }$, $\hat{\varphi }:E^{\prime }\to E$ with $\hat{\varphi }\varphi =[2]$. Compute the torsion subgroup using reduction of torsion points on an elliptic curve at several good primes, followed by division polynomials of an elliptic curve to identify the possible points.

For the free part, use two-isogeny descent. The two-torsion square-class homomorphism isin ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$, with kernel $\hat{\varphi }{E}^{\prime }(\mathbb{Q})$; define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly. The prime-support bound in two-isogeny descent restricts the first image to signed square-free divisors of $b$ and the second to those of $a^2-4b$. For each candidate $d$, solve its two-isogeny descent quarticTest local solubility at the real place and the relevant p-adic fields, discard impossible classes, and search the surviving coverings for rational points. A point with $UV\ne 0$ gives $x=d(U/V{)}^2$, $y=dUN/V^3$. Include the identity and rational 2-torsion cases separately. Once both rational images are determined, the two-isogeny rank formula givesThe point searches give explicit candidates for a basis. Their canonical height pairing can certify independence; Mordell-Weil saturation and effective height bounds then certify that they generate the full free part. Rank alone does not certify generators of the whole Mordell-Weil group. If the rational images are not fully determined, the procedure instead supplies lower and upper bounds on the rank of an elliptic curve.

We use the canonical height pairing normalization $⟨P,Q⟩=(\hat{h}(P+Q)-\hat{h}(P)-\hat{h}(Q))/2$, so that $⟨P,P⟩=\hat{h}(P)$. The supplied values giveThe convention without the factor $1/2$ instead gives the doubled pairing $B(P,Q)\approx 7.50$; all independence conclusions are the same.

In our convention the Gram matrix is approximatelyThis positive margin is much larger than the rounding uncertainty. Even allowing each of the three stated heights an error of $0.01$, we have $a\ge 6.60$, $b\ge 7.86$ and $|c|\le 3.765$, giving $ab-c^2\ge 6.60\cdot 7.86-3.765^2=37.700775>0$. Thus the actual canonical height pairing matrix is positive-definite, not merely its rounded approximation.

If $rP+sQ=O$ for integers $r,s$, bilinearity and the height parallelogram identity give $0=\hat{h}(rP+sQ)=a{r}^2+2crs+b{s}^2$. Positive definiteness forces $r=s=0$. This is the rank certificate from rounded canonical heights, and provesHere the last brackets denote the generated subgroup, not the scalar height pairing. The argument proves independence; it does not assert that these points generate the entire Mordell-Weil group.

First compute the Mordell-Weil group rank by two-isogeny descent. HereThe two-isogeny formula gives $\varphi :E\to E^{\prime }$. If $u=1+i$, then $u^4=-4$, andis an isomorphism over $K$. Its $x$-coordinate multiplier $u^2$ is a square in $K$.

Use the two-torsion square-class homomorphismFor any prime ideal of the Gaussian integers, if $v(x)>0$, then $x^2+i$ is a unit and $2v(y)=v(x)$; if $v(x)<0$, the term $x^3$ has strictly smallest valuation and $2v(y)=3v(x)$. Hence every valuation of $x$ is even. Since the Gaussian integers form a principal ideal domain, dividing by a square leaves a unit. Their units are $\{1,-1,i,-i\}$, whose square classes are $\{1,i\}$ because $-1=i^2$ and $i$ is not a square in $K$. For the latter assertion, $(a+bi{)}^2=i$ with $a,b\in \mathbb{Q}$ would imply $a^2=b^2$ and $2ab=1$, hence $a^2=1/2$, impossible for rational $a$. ThusBoth classes occur, at $O$ and $T$. This is the unit square-class bound for two-isogeny descent.

Define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly, with ${\alpha }^{\prime }((0,0))=[-4i]=[i]$. Since $\theta$ multiplies nonexceptional $x$-coordinates by a square, and preserves the exceptional classes as well, its image is also $\{1,[i]\}$. The standard kernel identities in two-isogeny descent areHere $\hat{\varphi }$ is the dual isogeny and $\hat{\varphi }\varphi =[2]$. These identities can be checked directly from the formulas: $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$, and conversely a square $x$-coordinate lets the quadratic equation for a preimage be solved using the curve equation. For example, if $x=r^2\ne 0$, the equation for $X$ is $X^2-(4x+2a)X+(a^2-4b)=0$, whose discriminant is $16(x^2+ax+b)=16(y/r{)}^2$; the $Y$-coordinate then follows from the dual formula. The exceptional points satisfy the same completed square-class criterion.

The two-isogeny index formula over a number field keeps track of a small kernel factor:In this case $\ker \hat{\varphi }=\{O,T^{\prime }\}$, where $T^{\prime }=(0,0)$, and $T^{\prime }\notin \varphi E(K)$ because ${\alpha }^{\prime }(T^{\prime })=[i]\ne 1$. Thus $\delta =2$, and the index is $2\cdot 2/2=2$. There is only one nonzero rational 2-torsion point on $E$: the other two would require $x^2=-i$, and $-i$ has the same nonsquare class as $i$. The Mordell-Weil theorem now givesso $r=0$.

It remains to identify all torsion, rather than merely the rank. The elliptic-curve discriminant is $-64i^3$, so the primes $(3)$ and $(2-i)$ have good reduction, with residue characteristics three and five. By the supplied point-count information their reduction groups have orders that are powers of two. The reduction of torsion points on an elliptic curve is injective on prime-to-residue-characteristic torsion. Every odd-primary torsion subgroup therefore injects into a group of two-power order at at least one of these two primes, and must be zero. All torsion is two-primary.

Finally, if a point had order four, its double would be $T$. The elliptic-curve addition formula givesFor $2P=T$ the denominator is nonzero, so $x^2=i$, impossible in $K$. A point of higher two-power order would have a multiple of order four, so it too is excluded. Thus the only torsion points are $O,T$. Together with rank zero,

For $E:y^2=x^3+a{x}^2+bx$ with rational 2-torsion $(0,0)$, the quotient by that point is the two-isogenous curveThe two-isogeny descent maps a nonexceptional point to the square class of its $x$-coordinate, with $(0,0)$ mapping to $[b]$. The images are finite collections of squarefree divisors of $b$ and $a^2-4b$, determined by testing the associated homogeneous quartics for rational points. If their orders are $2^s$ and $2^{s^{\prime }}$, thenwhich determines the Mordell-Weil rank $r$.

The method requires a rational 2-isogeny, and deciding whether every locally soluble quartic is globally soluble can be difficult. Computing only local conditions gives a 2-isogeny Selmer group and hence an upper bound; a nontrivial Tate-Shafarevich group can make that bound strict. Even after finding the rank, a separate saturation and point search may be needed to find generators.

Because the canonical height of an elliptic curve is a quadratic form, polarization makesa symmetric bilinear form on the free part of the Mordell-Weil group. If $P_i^{\prime }={\sum }_j{U}_{ij}{P}_j$ is another integral basis, then $U\in {\mathrm{GL}}_r(\mathbb{Z})$ and the Gram matrices satisfySince $\det U=\pm 1$, their determinants agree. Thus the regulator of an elliptic curve is independent of the chosen basis.

Now let $Q_1,\dots ,Q_r$ be a basis for the free part of $E^{\prime }(\mathbb{Q})$. The images $\varphi (P_i)$ span a finite-index sublattice, so modulo torsionfor an integral matrix $A$ with nonzero determinant. The height identity givesTaking determinants in the two descriptions of this Gram matrix yieldsTherefore the required formula holds with $d=(\det A{)}^{-1}\in {\mathbb{Q}}^{\times }$.

The Gram matrix of the canonical height pairing of rational points gives a robust independence certificate. For two points, set $a=\hat{h}(P)$, $b=\hat{h}(Q)$ and $c=(\hat{h}(P+Q)-a-b)/2$. If bounds from rounded measurements imply $a>0$ and $ab-c^2>0$, the matrix is positive-definite. Any integer relation $rP+sQ=O$ would give $a{r}^2+2crs+b{s}^2=0$, forcing $r=s=0$. Thus the points generate a free abelian group of rank two. Checking an interval lower bound for the determinant, rather than just its approximate central value, makes the certificate insensitive to rounding. It certifies independence, not that the points are a basis of the entire Mordell-Weil group.

When an elliptic curve has full rational 2-torsion, its Kummer map embeds $E(\mathbb{Q})/2E(\mathbb{Q})$ into a product of square-class groups. Valuation and local-solubility conditions reduce the image to finitely many classes and thereby bound the rank of the Mordell-Weil group.